INTEGERS 11 (2011)
#A68
SUM-PRODUCT ESTIMATES APPLIED TO WARING’S PROBLEM
OVER FINITE FIELDS
Todd Cochrane1
Department of Mathematics, Kansas State University, Manhattan, Kansas
cochrane@math.ksu.edu
James Cipra
Department of Mathematics, Kansas State University, Manhattan, Kansas
cipra@math.ksu.edu
Received: 5/13/11, Accepted: 10/31/11, Published: 11/2/11
Abstract
Let A be the set of nonzero k-th powers in Fq and γ ∗ (k, q) denote the minimal n such
that nA = Fq . We use sum-product estimates for |nA| and |nA − nA|, following the
method of Glibichuk and Konyagin to estimate γ ∗ (k, q). In particular, we obtain
γ ∗ (k, q) ≤ 633(2k)log 4/ log |A| for |A| > 1 provided that γ ∗ (k, q) exists.
1. Introduction
Let Fq be a finite field in q = pf elements and k be a positive integer. The smallest
s such that the equation
xk1 + xk2 + · · · + xks = a
(1)
is solvable for all a ∈ Fq (should such an s exist) is called Waring’s number for Fq ,
denoted γ(k, q). Similarly, the smallest s such that
±xk1 ± xk2 ± · · · ± xks = a,
(2)
is solvable for all a ∈ Fq is denoted δ(k, p). If d = (k, q − 1) then clearly γ(d, q) =
γ(k, q) and so we may assume k|(q − 1). If A is the multiplicative subgroup of k-th
powers in F∗q then we write
γ(A, q) = γ(k, q),
δ(A, q) = δ(k, q).
Also, we let γ ∗ (A, q), δ ∗ (A, q) denote the minimal s such that every element of Fq
is the sum (± sum ) of exactly s nonzero k-th powers, that is, (1), (2) resp. are
1 Current
address: Dept. of Mathematics, Georgia Institute of Technology, Atlanta, Georgia
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solvable with all xi $= 0. It is well-known that γ(A, q), δ(A, q), γ ∗ (A, q), δ ∗ (A, q)
exist if and only if A contains a set of f linearly independent points over Fp ; see
Lemma 6.
For any subsets S, T of Fq and positive integer n, let
S + T = {s + t : s ∈ S, t ∈ T },
nS = S+S+· · ·+S (n-times),
S − T = {s − t : s ∈ S, t ∈ T },
ST = {st : s ∈ S, t ∈ T },
S n = SS · · · S (n-times).
Note that (nS)T ⊆ n(ST ). We let nST denote the latter, n(ST ). Also, for any
a ∈ Fq we let aS = {as : s ∈ S}.
If A is a multiplicative subgroup of F∗q then γ ∗ (A, q) (if it exists) is the minimal
s such that sA = Fq , while γ(A, q) is the minimal s such that sA0 = Fq , where
A0 = A∪{0}. It is well-known that γ(k, q) ≤ k with equality if k = q−1 or (q−1)/2.
This was first observed by Cauchy [5] for the case q = p. Our first result is the
analogue for γ ∗ (k, q). The proof makes use of Kneser’s lower bound for |A + B|.
Theorem 1. If A is the multiplicative subgroup of k-th powers in F∗q , |A| > 2 and
γ ∗ (A, q) exists, then γ ∗ (A, q) ≤ k + 1. When |A| = 2 (that is, q is an odd prime
and A = {±1}) then γ ∗ (A, q) = 2k.
For |A| > 2 it was established by Tietäväinen [20], for odd q, and by Winterhof
[22], [23, Lemma 1], for even q, that γ(A, q) ≤ [k/2] + 1. It is an open question
whether the same improvement holds for γ ∗ (A, q). For the case of prime fields
Heilbronn [17] formulated two conjectures, which in the more general setting of Fq
can be stated as follow:
√
1. If |A| > 2 then γ(A, q) ' k.
2. For any # > 0 there exists a constant c(#) such that if |A| > c(#) then
γ(A, q) '! k! .
The second conjecture was proven by Konyagin [18] for prime fields. Cipra,
Cochrane and Pinner [8] established
the first conjecture for prime fields, and the
√
explicit bound γ(A, p) ≤ 83 k was obtained in [9]. Cipra [6, Theorem 4] proved
the first conjecture for the general finite field Fq , obtaining
! √
16 k + 1, for q = p2 .
√
γ(A, q) ≤
(3)
10 k + 1, for q = pf , f ≥ 3,
whenever γ(A, q) is defined.
Next, let A" = A ∩ Fp , so that |A" | = (|A|, p − 1). Cipra [6], sharpening the work
of Winterhof [22], established the bound
"
#
(k + 1)1/f − 1
γ(k, q) ≤ 8f
,
(4)
|A" |
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whenever γ(k, q) exists. He also established the bound
log 4
γ(k, q) ' f k f log |A! | log log(k),
(5)
which resolved the second Heilbronn conjecture provided |A" |f is sufficiently large.
For prime fields Glibichuk and Konyagin [15] used methods of additive combinatorics to obtain
log 4
γ ∗ (A, p) ≤ 400 k log |A| ,
(6)
for any multiplicative subgroup A with |A| > 1. Cochrane and Pinner [9, Corollary
7.1] obtained a similar bound for q = p, and Glibichuk [13] established the same
type of bound for q = p2 . The main result of this paper is a generalization of (6)
to arbitrary Fq , thus resolving the second Heilbronn conjecture for any finite field.
Theorem 2. If A is a multiplicative subgroup of F∗q for which γ ∗ (A, q) is defined
and |A| > 1, then with k = (q − 1)/|A|, we have
log 4
γ ∗ (A, q) ≤ 633(2k) log |A| .
After submitting this work, the author’s learned that Glibichuk [14, Corollary 1]
recently proved a similar result, albeit with weaker constants. In particular, if |A| =
p! , then our result gives γ ∗ (A, q) ' 41/! , whereas his result gives γ ∗ (A, q) ' 61/! .
For δ ∗ (A, q) we establish the stronger bound,
Theorem 3. If A is a multiplicative subgroup of F∗q for which δ ∗ (A, q) is defined
and |A| > 1, then with k = (q − 1)/|A|, we have
log 4
δ ∗ (A, q) ≤ (40/3)k log |A| .
As noted in Theorem 9, if q is even or |A| is even then γ ∗ (A, q) = δ ∗ (A, q) and
thus the stronger bound in Theorem 3 applies to γ ∗ (A, q) as well. Further relations
between δ(A, q) and γ(A, q) are given in Theorem 9. The exponent on k in the
theorem improves on (5) when |A| > (|A|, p − 1)f and on (4) when |A| > 4f . For
small |A| (|A| = O(1) as p → ∞) one can obtain a stronger result by employing the
lattice method of Bovey. In this manner we prove,
Theorem 4. For any positive integer t there is a constant c1 (t) such that if A is a
multiplicative subgroup of F∗q with |A| = t, and such that γ(A, q) is defined, then
γ(A, q) ≤ c1 (t)k1/φ(t) .
The constant c1 (t), estimated in [4] for prime fields, depends on the size of the
coefficients of the cyclotomic polynomial of order t.
Corollary 5. For any positive integer l there is a constant c(l) such that if A is a
multiplicative subgroup of F∗q of order t such that φ(t) ≥ l and γ(A, q) exists, then
γ(A, q) ≤ c(l)k1/l .
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Proof. Suppose that φ(t) ≥ l. Put c = maxt≤4l c1 (t), c(l) = max{c, 21/l 633}, with
c1 (t) as defined in Theorem 4. If t > 4l then, by Theorem 2, γ(k, q) ≤ 633·21/l k1/l ≤
c(l)k1/l . If t ≤ 4l then, by Theorem 4, γ(k, q) ≤ ck1/φ(t) ≤ c(l)k1/l .
!
2. Preliminary Lemmas
The first lemma gives equivalent conditions for the existence of γ(k, q). It is wellknown, and follows from the fact that the set of all sums of k-th powers is a multiplicatively closed set and therefore a subfield of Fq ; see Tornheim [21, Lemma 1],
or Bhaskaran [1].
Lemma 6. Let A be the set of nonzero k-th powers in Fq . The following are
equivalent.
(i) γ(k, q) exists, that is, every element of Fq is a sum of k-th powers.
(ii) A is not contained in any proper subfield of Fq ; that is, A contains a set of f
linearly independent points over Fp .
(iii) |A| does not divide pj − 1 for any j|f , j < f , that is,
for any j|f , j < f .
pf −1
pj −1
does not divide k
It is also not hard to show that γ ∗ (A, q) exists if and only if |A| > 1 and γ(A, q)
exists.
An important tool needed throughout this paper is Rusza’s triangle inequality
(see, e.g., Nathanson [19, Lemma 7.4]),
|S + T | ≥ |S|1/2 |T − T |1/2 ,
(7)
for any S, T ⊆ Fq , and its corollary
1
1
1
|nS| ≥ |S| 2n−1 |S − S|1− 2n−1 ≥ |S − S|1− 2n ,
(8)
for any positive integer n. The first inequality in (8) follows by induction on n, and
the second from the trivial bound |S − S| ≤ |S|2 .
The following is a key lemma for showing that a sum-product set fills up Fq .
Lemma 7. Let A, B be subsets of Fq and m ≥ 3 be a positive integer.
$
2
(a) If |B||A|1− m > q 1 −
|B|
q
%2/m
then mAB = Fq .
(b) If |B||A| ≥ 2q then 8AB = Fq .
(c) If |B||A| > q and either A or B is symmetric (A = −A) or antisymmetric
(A ∩ −A = ∅) then 8AB = Fq .
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A slightly weaker form of part (a) was proven by Bourgain [2, Lemma 1] for q = p
and m = 3 and by Cochrane and Pinner [9, Lemma 2.1] for q = p and general m. A
similar proof works for Fq and is provided in Section 7. In the earlier versions of this
$
%2/m
statement, an extra hypothesis, 0 $∈ A, was included, and the factor 1 − |B|
q
was excluded.
Part (b) is due to Glibichuk and Konyagin [15, Lemma 2.1] for prime fields and
to Glibichuk and Rudnev [16] for general Fq . Part (c) is due to Glibichuk [12] for
prime fields and to Glibichuk and Rudnev [16] as well as Cipra [7] for general Fq .
In particular, if A is a multiplicative subgroup, then applying (c) with B = A we
√
see that γ ∗ (A, q) ≤ 8 provided that |A| > q.
In the cases where |A| = 3, 4 or 6 one can actually evaluate γ(A, q). This was
done in [8] for the case of prime moduli.
Theorem 8. Let A is a multiplicative subgroup of Fq of order 3,4 or 6 for which
γ(A, q) exists. Then q = p or p2 . If q = p2 then γ(A, q) = p − 1. If q = p then


if |A| = 3,
a + b − 1,
γ(A, p) = c − 1,
if |A| = 4,

+
* 2
1
a
+
b
,
if |A| = 6
3
3
where, if |A| = 3 or 6, then a, b are the unique positive integers with a > b and
a2 + b2 + ab = p, while if |A| = 4 then c, d are the unique positive integers with c > d
and c2 + d2 = p.
In particular, for |A| = 3, 4 or 6 we have
√
√
3k + 1 − 1 ≤γ(A, q) ≤ 2 k,
√
√
2k − 1 ≤γ(A, q) ≤ 2 k − 1,
√
√
2k − 12 ≤γ(A, q) ≤ 23 6k,
if |A| = 3,
if |A| = 4,
if |A| = 6.
Proof. Since |A| = 3, 4 or 6, every element of A is of degree 1 or 2 over Fp and
therefore A ⊂ Fp2 . Thus, in order for γ(k, q) to exist we must have q = p or p2 .
The case q = p is just Theorem 2 of [8] and the case q = 22 is trivial, so we shall
assume q = p2 with p an odd prime and that A is not contained in Fp .
Case i: |A| = 3. Say A = {1, T, T 2 } where T ∈ Fp2 − Fp satisfies T 2 + T + 1 = 0.
In particular, p ≡ 2 (mod 3).
√ We claim that γ(A, q) = p − 1 and consequently,
since 3k = p2 − 1, γ(k, q) = 3k + 1 − 1. Let w = x + yT denote a typical element
of Fq where 0 ≤ x, y ≤ p − 1 and let γ(w) denote the minimal number of elements
of A required to represent w. First note that γ(0) = 3 since 1 + T + T 2 = 0
so we assume that w $= 0. Suppose that x ≤ y. If x + y < p then trivially
γ(w) < p. If x ≤ y < 2x then we write w = (y − x)T + (p − x)T 2 and get
γ(w) ≤ p+y−2x < p. If y ≥ 2x and y > 23 p then we write w = (x−y+p)·1+(p−y)T 2
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and get γ(w) ≤ (x − 2y + 2p) ≤ 2p − 32 y < p. If y ≥ 2x and y < 23 p, then
x + y ≤ 32 y < p. A similar argument holds for x ≥ y. Finally, one can check that
γ( 13 (p + 1) + 23 (p + 1)T ) = p − 1.
Case ii: |A| = 4. Say A = {±1, ±T }, with T 2 = −1, T ∈ Fp2 − Fp . In particular,
p ≡ 3 (mod 4). Any element of Fq may be written x+yT with |x|, |y| ≤ p−1
2 , and so
p−1
γ(A, q) ≤ p − 1. Also, it is plain that γ( p−1
+
T
)
=
p
−
1.
Thus,
γ(A,
q)
= p − 1.
2
2
Case iii: |A| = 6. Say A = {±1, ±T, ±T 2 } with T 2 − T + 1 = 0. As in case ii,
any element of Fq may be written x + yT with |x|, |y| ≤ p−1
2 , and so γ(A, q) ≤ p − 1.
p−1
Also, with just a little work one again sees that γ( 2 + p−1
2 T ) = p − 1. Thus,
γ(A, q) = p − 1.
!
The precise relationship between γ(k, q) and δ(k, q) is an important unresolved
problem. It is not known whether γ(k, q) ≤ Cδ(k, q) for some constant C. Bovey [4,
Lemma 2] established γ(k, p) ≤ (log2 p + 1)δ(k, p) for prime moduli, and improvements were given in [8]. Here we prove the analogue of [8, Theorem 4.1] for general
finite fields.
Theorem 9. Let A be the set of nonzero k-th powers in Fq with k|(q − 1), such
that γ(k, q) is defined. Then,
,
$
%3 log q
(a) γ(k, q) ≤ 3 log2 log
δ(k, q).
|A|
(b) γ(k, q) ≤ 3 0log2 (4δ(k, q))1 δ(k, q)
(c) γ(k, q) ≤ 2 0log2 (log2 (q))1 δ(k, q).
(d) γ(k, q) ≤ (pmin − 1)δ(k, q), where pmin is the minimal prime divisor of |A|.
(e) If q is even or |A| is even then δ(k, q) = γ(k, q). If |A| is odd and p is odd,
then δ(k, q) = γ( k2 , q).
Proof. a) Put A0 = A ∪ {0}, δ = δ(k, q). Since δA0 − δA0 = Fq we obtain from (8),
(observing that this inequality is strict for |S| > 1),
j
j
|jδA0 | > |δA0 − δA0 |1−1/2 = q 1−1/2 ,
(9)
$
%
1
3 log q
for any positive integer j. Hence if j ≥ log2 log
we have |jδA0 ||A| 3 ≥ q, and
|A|
so by Lemma 7 (a) with m = 3, 3(jδA0 )A = Fq , that is, 3jδA0 = Fq .
b) This follows from part (a) and the trivial bound (2|A|+1)δ ≥ q, when |A| ≥ 11.
Indeed, in this case,
log q
log(2|A| + 1)
4
≤δ
< δ.
log |A|
log |A|
3
For |A| < 11, the result follow from part (d) of this theorem, since pmin ≤ 7 for
such |A|.
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INTEGERS: 11 (2011)
j
c) We repeat the proof given by Cipra [6]. If j ≥ log2 (log2 (q)) then q 1/2 ≤ 2
and so by (9), |jδA| > q/2. Thus, 2jδA = Fq . (Here we have used the fact that if
S is a subset of a finite group G with |S| > |G|/2 then S + S = G.)
d) Let % be the minimal prime divisor of |A|. Then A has a subgroup G of order
.
% and x∈G x = 0 so that −1 is a sum of % − 1 elements of A.
e) If q is even then 1 = −1, and so trivially δ(k, q) = γ(k, q). If |A| is even then
−1 is a k-th power, and so again γ(k, q) = δ(k, q). If |A| is odd then k must be even
(for p $= 2) and A ∪ (−A) is the set of k/2-th powers.
!
3. Proof of Theorem 1
Let k|(q − 1), A be the set of nonzero k-th powers in Fq and A0 = A ∪ {0}.
Before addressing Theorem 1, which is concerned with representing elements as
sums of nonzero k-th powers, we start by reviewing the proof of Cauchy’s theorem,
γ(k, q) ≤ k, which allows for some terms to be zero. For any positive integer n,
nA0 = {0} ∪ Ax1 · · · ∪ Axl ,
for some distinct cosets Axi of A, 1 ≤ i ≤ l. If nA0 $= Fq then (n+1)A0 contains nA0
and, assuming that γ(k, q) exists, must be strictly larger. Therefore, |(n + 1)A0 | ≥
|nA0 | + |A|. By induction we get a Cauchy-Davenport type inequality,
|nA0 | ≥ min{q, 1 + n|A|},
(10)
for n ≥ 1, and in view of the equality k|A| = q −1, deduce that γ(k, q) ≤ k whenever
γ(k, q) exists. To estimate γ ∗ (k, q) we have to work a little harder since (n + 1)A
doesn’t contain nA in general, so it is not immediate that it has larger cardinality.
However, we are able to recover the following analogue of (10), and Theorem 1 is
an immediate consequence.
Lemma 10. If A is a multiplicative subgroup of F∗q containing f linearly independent points over Fp and |A| > 2, then for any positive integer n, |nA| ≥
min{q, n|A|}.
Proof. Let A be a multiplicative subgroup of Fq containing f linearly independent
points over Fp . We first show that if B is any subset of Fq such that AB ⊂ B
then either A + B = Fq or |A + B| ≥ |A| + |B| − 1. This follows from Kneser’s
inequality (see [19, Theorem 4.1]): |A + B| ≥ |A| + |B| − |stab(A + B)|, where
stab(A + B) = {x ∈ Fq : A + B + x = A + B}, an additive subgroup of Fq , that
is, an Fp subspace of Fq . We need only establish that if stab(A + B) $= {0} then
A+B = Fq . Suppose x is a nonzero element of stab(A+B). Then A+B+x = A+B.
Since AB ⊂ B and AA = A we get A+B+Ax ⊂ A+B. Thus Ax ⊂ stab(A+B), but
INTEGERS: 11 (2011)
8
Ax contains f linearly independent points over Fp . Thus stab(A+B) is of dimension
f over Fp , and so stab(A + B) = Fq . Plainly, we must also have A + B = Fq since
for any point c ∈ A + B, c + stab(A + B) ⊂ A + B.
Now let n be any positive integer and B = nA. Then AB ⊂ B and so either
(n + 1)A = Fq or |(n + 1)A| ≥ |nA| + |A| − 1. The proof now follows by induction on
n. Suppose |nA| ≥ n|A| for a given n and that (n + 1)A $= Fq . Then |(n + 1)A| ≥
(n + 1)|A| − 1, but (n + 1)A is a union of cosets of A together (possibly) with 0. If
|A| > 2, this forces (n + 1)A to be a union of at least (n + 1) cosets of A together
(possibly) with 0. Thus |(n + 1)A| ≥ (n + 1)|A|.
!
When |A| = 2; that is, q = p and A = {±1}, then |nA| = n + 1 for n < p. Thus
γ ∗ (A, q) = p − 1 = 2k.
4. Estimating δ ∗ (A, q) and Proof of Theorem 3
Following the method of Glibichuk and Konyagin [15], for any subsets X, Y of Fq
let
/
0
X −X
x1 − x2
=
: x1 , x2 ∈ X, y1 , y2 ∈ Y, y1 $= y2 .
Y −Y
y1 − y2
The key lemma is a generalization of a lemma of Glibichuk and Konyagin [15,
Lemma 3.2] to finite fields.
Lemma 11. Let q = pf , X, Y ⊆ Fq and a1 , a2 , . . . , af ∈ Fq be a set of f linearly
independent points over Fp . If X−X
Y −Y $= Fq then for some ai we have
|2XY − 2XY + ai Y 2 − ai Y 2 | ≥ |X||Y |.
Proof. Let S = X−X
Y −Y . Assume S $= Fq and that a1 , . . . , af are linearly independent
values in Fq . We claim that for some ai , S + ai $⊆ S, for otherwise S + k1 a1 + k2 a2 +
· · · + kf af ⊆ S for all nonnegative integers k1 , . . . , kf , implying that S = Fq . Say
x1 −x2
/ S, for some x1 , x2 ∈ X, y1 , y2 ∈ Y . Then the mapping from X × Y
y1 −y2 + ai ∈
into 2XY − 2XY + ai Y 2 − ai Y 2 given by
(x, y) → (y1 − y2 )x + (x1 − x2 + ai y1 − ai y2 )y,
is one-to-one and the lemma follows.
!
Applying the lemma to a multiplicative subgroup A of F∗q containing a set
a1 , . . . , af of linearly independent points, we immediately obtain,
Lemma 12. Let A be a multiplicative subgroup of F∗q containing f linearly independent points over Fp and X be any subset of Fq such that AX ⊆ X and X−X
A−A $= Fq .
Then
|2X − 2X + A − A| ≥ |X||A|.
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We also need the following elementary result.
Lemma 13. Let A be a multiplicative subgroup of F∗q and X, Y be subsets of Fq
such that AX ⊆ X, AY ⊆ Y . If |X − X||Y − Y | ≤ q|A| then X−X
Y −Y $= Fq .
Proof. If c = (x1 − x2 )/(y1 − y2 ) for some x1 , x2 ∈ X, y1 $= y2 ∈ Y , then c =
(ax1 − ax2 )/(ay1 − ay2 ) for any a ∈ A. Thus
1
1
1 X − X 1 |X − X|(|Y − Y | − 1)
1
1
.
1Y −Y 1≤
|A|
Since the right-hand side is less than q by assumption, the result follow.
!
5 l
For l ∈ N, let n1 = 1 and nl = 24
4 − 13 , for l ≥ 2, so that n2 = 3, n3 = 13,
n4 = 53, n5 = 213 and
nl+1 = 4nl + 1, for l ≥ 2.
(11)
Put A1 = A and, for l ≥ 2, Al = (nl A − nl A) so that, for l ≥ 2,
2Al−1 − 2Al−1 + A − A = Al .
(12)
Lemma 14. Let A be a multiplicative subgroup of F∗q containing f linearly independent points over Fp . Then for l ≥ 1,
(a) If
|Al−1 − Al−1 ||A − A| < q|A| then |Al | ≥ |A|l .
(b) In all cases, |Al | ≥ min{|A|l , q/|A|}.
One can compare the above result with Lemma 5.2 of [15] where it is shown for Fp
that |Al | ≥ 38 min{|A|l , p−1
2 }.
Proof. The proof of (a) is by induction on l, the statement being trivial for l = 1.
For l > 1, put X = Al−1 , Y = A. If |Al−1 − Al−1 ||A − A| < q|A| then by
Lemma 13, X−X
Y −Y $= Fq . Also, by (12) we have 2X − 2X + A − A = Al . Thus by
Lemma 12, |Al | ≥ |Al−1 ||A| and so by the induction assumption, |Al | ≥ |A|l . If
|Al−1 − Al−1 ||A − A| ≥ q|A| then since |A − A| ≤ |A|2 we have |Al−1 − Al−1 | ≥
q/|A|. Since |Al | = |nl A − nl A| ≥ |2nl−1 A − 2nl−1 A| = |Al−1 − Al−1 |, we obtain
|Al | ≥ |Al−1 − Al−1 | ≥ q|A|/|A − A| ≥ q/|A|.
!
Lemma 15. Let A be a multiplicative subgroup of F∗q containing f linearly independent points over Fp . Set l = 3log(q − 1)/ log |A|4. Then δ ∗ (A, q) ≤ 16nl .
Proof. For such l we have l + 1 > log(q − 1)/ log |A| and so |A|l+1 ≥ q. Thus, by
Lemma 14 (b), |Al ||A| ≥ min{|A|l+1 , q} = q. Since (|A|, q) = 1 we must in fact have
|Al ||A| > q. Since A is symmetric (−1 ∈ A) or antisymmetric (−1 ∈
/ A), it follows
from Lemma 7 (c) that 8Al A = Fq , that is, 8nl A − 8nl A = Fq .
!
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Proof of Theorem 3. With l as in Lemma 15 we have using k|A| = (q − 1),
2
3
log(q − 1)
log k
l=
≤1+
.
log |A|
log |A|
5 l
Thus by Lemma 15, δ ∗ (A, q) ≤ 16nl ≤ 16 24
4 ≤
the proof.
40 log k/ log |A|
,
3 4
thereby finishing
!
5. Estimating γ ∗ (A, q) and Proof of Theorem 2
Let A be a multiplicative subgroup of F∗q containing f linearly independent points.
We start by obtaining growth estimates for |nA|. If |A| = 1 then q = p and |nA| = 1
for any n. If |A| = 2 then q = p, A = ±1 and |nA| = min{p, n + 1}. Next we note
that
!
|A|3/2
if
|A − A|2 < q|A|,
|4A| ≥
(13)
q 1/2 |A|3/8 , otherwise.
Indeed, by (7) and Lemma 14 (a) we have
|4A| ≥ |A|1/2 |3A − 3A|1/2 = |A|1/2 |A2 |1/2 ≥ |A|3/2 ,
if
|A − A|2 < q|A|.
Otherwise, |A − A| ≥ (q|A|)1/2 . In particular, |A|2 > (q|A|)1/2 , and so |A| ≥ q 1/3 .
Thus, by (8), |4A| ≥ |A−A|15/16 ≥ (q|A|)15/32 ≥ q 15/32 |A|3/32 |A|12/32 ≥ q 1/2 |A|3/8 .
For l ∈ N set
5 l l
2
ml =
4 − + ,
18
3 9
so that m1 = 1, m2 = 4, m3 = 17, m4 = 70, m5 = 283 and ml = ml−1 + nl for
l ≥ 2, with nl as defined in the previous section.
Lemma 16. Let A be a multiplicative subgroup of F∗q containing f linearly independent points over Fp . Then for l ≥ 1 we have
 l−1+ 1
|A| / 2l−1 , if l = 1, or l ≥ 2 and
0
$
%3/4
|ml A| ≥
|A|
3/4
1/4
1/2
max
q , |A| q
,
|A−A|
|Al−1 − Al−1 ||A − A| < q|A|,
otherwise.
(14)
Proof. The result is trivial when l = 1. Assume that the theorem holds for l − 1
with l ≥ 2. Suppose that |Al−1 − Al−1 ||A − A| < q|A|. In particular, if l ≥ 3
then |Al−2 − Al−2 ||A − A| < q|A|. Then using ml = nl + ml−1 , inequality (7), the
induction assumption and Lemma 14, we have
1
1
1
|ml A| ≥ |ml−1 A|1/2 |nl A − nl A|1/2 ≥ |A|(l−2+ 2l−2 ) 2 |A|l/2 = |A|l−1+ 2l−1 .
11
INTEGERS: 11 (2011)
Suppose next that
(15)
|Al−1 − Al−1 ||A − A| ≥ q|A|.
Then, since ml > nl > 4nl−1 , we have by inequality (8) that
|ml A| ≥ |2(2nl−1 A)| ≥ |2nl−1 A − 2nl−1 A|3/4 = |Al−1 − Al−1 |3/4 ≥
4
q|A|
|A − A|
53/4
.
To prove the second inequality, |ml A| ≥ |A|1/4 q 1/2 , under the assumption of (15),
more work is required. A stronger result was established (13) for l = 2, so we assume
l ≥ 3. First observe that by Lemma 12, with X = A − A, if |2A − 2A||A − A| < q|A|,
(so that by Lemma 13, (X − X)/(A − A) $= Fq ), then
|5A − 5A| = |2(A − A)A − 2(A − A)A + A − A| ≥ |A − A||A|,
and so by (15),
|5A−5A||2nl−1 A−2nl−1 A| = |5A−5A||Al−1 −Al−1 | ≥ |A−A||A||Al−1 −Al−1 | ≥ q|A|2 .
Since 2nl−1 > 5 for l ≥ 3, it follows that
|2nl−1 A − 2nl−1 A| > |5A − 5A|1/2 |2nl−1 A − 2nl−1 A|1/2 ≥ q 1/2 |A|.
(16)
Also, since for any set B, |B − B| ≤ |B|2 , using (15),
|2nl−1 A|2 |A|2 ≥ |2nl−1 A − 2nl−1 A||A − A| = |Al−1 − Al−1 ||A − A| ≥ q|A|,
and so
|2nl−1 A|2 |A| ≥ q.
(17)
Thus since ml > nl > 4nl−1 we obtain from (7), (16) and (17),
|ml A| ≥ |4nl−1 A| ≥ |2nl−1 A|1/2 |2nl−1 A − 2nl−1 A|1/2 ≥ |2nl−1 A|1/2 q 1/4 |A|1/2
= (|2nl−1 A|1/2 |A|1/4 )q 1/4 |A|1/4 ≥ q 1/4 q 1/4 |A|1/4 = q 1/2 |A|1/4 .
There remains the case |2A − 2A||A − A| ≥ q|A|. In this case |2A − 2A| ≥
q 1/2 |A|1/2 and |2A|2 |A| ≥ q. The latter implies |2A| ≥ (q/|A|)1/2 . Thus, by (7),
|4A| ≥ |2A|1/2 |2A − 2A|1/2 ≥ q 1/4 |A|−1/4 q 1/4 |A|1/4 = q 1/2 ,
and
|ml A| ≥ |6A| ≥ |4A|1/2 |2A − 2A|1/2 ≥ q 1/4 q 1/4 |A|1/4 = q 1/2 |A|1/4 ,
which finishes the proof.
!
Lemma 17. Let A be a multiplicative subgroup of F∗q containing f linearly independent points over Fp . Then for l ≥ 1 we have
6
7 8
1
|ml A| ≥ min |A|l−1+ 2l−1 , 2q .
(18)
12
INTEGERS: 11 (2011)
Proof. If l = 1 or |A| = 1 the statement is trivial. For |A| = 2,
7
5 l l
11
4 − + , q} ≥ min{2l , 2q}.
18
3
9
√
For |A| ≥ 4 the result follows from Lemma 16, since q 1/2 |A|1/4 ≥ 2q in this case.
For |A| = 3 and l = 2 the result follows from (13) in a similar manner. Suppose that
|A| = 3 and l ≥ 3. Then A = {1, α, α2 } = {1, α, −1−α}, where α is a primitive cube
root of 1, and A−A = {0, ±(1−α), ±(2+α), ±(2α+1)}, whence |A−A| = 7. Then,
√
by Lemma 16, |ml A| ≥ q 3/4 |A|3/4 |A − A|−3/4 ≥ (3/7)3/4 q 3/4 ≥ 2q, provided that
q ≥ 4(7/3)3 = 50.8.., and so the result follows from Lemma 16 when q ≥ 51. We
are left with testing the prime powers less than 50. If q is a prime then we can use
√
the Cauchy-Davenport inequality to get |mk A| ≥ |17A| ≥ min{q, 35} ≥ 2q. The
prime powers remaining with 3|q − 1 are 4,16,25,49. We can’t have q = 16 or 49
since 3|(22 − 1) and 3|(7 − 1), implying that A does not contain a set of f linearly
√
∗
independent points. For q = 4, A = F√
2q. For
4 , 2A = F4 and trivially |2A| ≥
q = 25 one can check that |3A| = 10 ≥ 50.
!
|ml A| ≥ min{ml + 1, q} = min{
Applying Lemma 7 (b) with A = B = ml A we immediately obtain,
Lemma 18. Let A be a multiplicative subgroup of F∗q containing f linearly inde−1
1
1
pendent points over F . If |A| ≥ (2q) 2 (l−1+ 2l−1 ) , then 8m2 A = F .
p
l
In particular,
8A = Fq
128A = Fq
2312A = Fq
39200A = Fq
640712A = Fq
for |A| > q 1/2 ,
for |A| > 1.26 q 1/3 ,
for |A| > 1.17 q 2/9 ,
for |A| > 1.12 q 4/25 ,
for |A| > 1.09 q 8/65 .
In comparison, for Fp Cochrane and Pinner [9] obtained
8A = Fp
32A = Fp
392A = Fp
2888A = Fp
12800A = Fp
56448A = Fp
228488A = Fp
for |A| > p1/2 ,
for |A| > 3.91 p1/3 ,
for |A| > 2.78 p1/4 ,
for |A| > 3.19 p1/5 ,
for |A| > 2.28 p1/6 ,
for |A| > 2.43 p1/7 ,
for |A| > 1.91 p1/8 .
q
13
INTEGERS: 11 (2011)
We cannot do quite as well for Fq because we do not have a lower bound on |2A|.
For Fp it was shown in [9, Theorem 5.2] that |2A| ≥ min{ 14 |A|3/2 , p2 }. We do not
know if such a bound holds in Fq .
Proof of Theorem 2. An integer l satisfies the hypothesis of Lemma 18 provided
that
1
log 2q
l + l−1 ≥
+ 1.
(19)
2
2 log |A|
We claim that the value
"
#
log(2(q − 1))
l=
+1 ,
2 log |A|
suffices. To see this, first observe that for this choice of l, l < log2 (4(q − 1)), that
is, q − 1 > 2l−2 , and so using the inequality
log(2q) − log(2(q − 1)) = log(q/(q − 1)) <
we have
q
1
−1=
,
q−1
q−1
log(2q)
log(2(q − 1))
1
1
1
−
<
< l−1
< l−1 .
2 log |A|
2 log |A|
2(q − 1) log |A|
2 log |A|
2
Thus (19) is satisfied, and so by Lemma 18, γ ∗ (A, q) ≤ 8m2l . Since ml ≤
have
γ ∗ (A, q) ≤ 8(5/18)2 420
5 l
18 4 ,
we
log 2(q−1)
+1
2 log |A|
log 4
1 < 158.03 · 4 loglog2(q−1)
|A|
= 158.03(2(q − 1))log 4/ log |A|
log 4
= 158.03(2|A|k) log |A| ≤ 633(2k) log |A| ,
completing the proof of Theorem 2.
!
We cannot do quite as well for Fq as for Fp because we do not have a good
lower bound on |2A|. For Fp we were able to use the bound [9, Theorem 5.2],
|2A| ≥ min{ 14 |A|3/2 , p2 }. We do not know if such a bound holds in Fq .
6. Proof of Theorem 4
For small |A| we use the method of Bovey [4] to bound δ(A, q) and γ(A, q). We
start by generalizing [4, Lemma 3]. For any n-tuple u = (u1 , . . . , un ) ∈ Rn let
.n
5u51 = i=1 |ui |.
Lemma 19. Let Fq be any finite field and u1 , u2 , . . . , un ∈ Fq . Let T : Zn → Fq
.n
n
be the linear function T (x1 , . . . , xn ) =
i=1 xi ui . Suppose that v1 , . . . , vn ∈ Z
are linearly independent vectors over R with T (vi ) = 0, 1 ≤ i ≤ n. Then for
any value a in the range of T there exists a vector u ∈ Zn with T (u) = a and
.n
5u51 ≤ 12 i=1 5vi 51 .
14
INTEGERS: 11 (2011)
.n
Proof. Let w ∈ Zn with T (w) = a. Write w = i=1 yi vi for some yi ∈ R, 1 ≤ i ≤ n.
Say yi = xi + #i for some xi ∈ Z and #i ∈ R with |#i | ≤ 1/2, 1 ≤ i ≤ n. Put u =
.n
.n
.n
1
n
!
i=1 #i vi = w −
i=1 xi vi . Then u ∈ Z , T (u) = a and 5u51 ≤ 2
i=1 5vi 51 .
Proof of Theorem 4. By Theorem 9 (d), γ(k, q) ≤ (t − 1)δ(k, q), where t = |A|, and
so it suffices to prove the statement of Theorem 4 for δ(k, q). Put r = φ(t). Let
R be a primitive t-th root of unity in Fq , Φt (x) be the t-th cyclotomic polynomial
over Q of degree r and ω be a primitive t-th root of unity over Q. In particular,
Φt (R) = 0 and the set of nonzero k-th powers in Fq is just {1, R, R2 , . . . , Rt−1 }.
Let f : Zr → Z[ω] be given by
f (x1 , x2 , . . . , xr ) = x1 + x2 ω + · · · + xr ω r−1 .
Then f is a one-to-one Z-module homomorphism.
.r
i−1
Let T : Zr → Fq be the linear map T (x1 , . . . , xr ) =
and L be
i=1 xi R
the lattice of points satisfying T (x1 , . . . , xr ) = 0. Since the set of k-th powers
1, R, . . . , Rr−1 spans all of Fq we have V ol(L) = q. Thus by Minkowski’s fundamental theorem there is a nonzero vector v1 = (a1 , a2 , . . . , ar ) in L with |ai | ≤ q 1/r ,
1 ≤ i ≤ r. For 2 ≤ i ≤ r set vi = f −1 (ω i−1 f (v1 )). Then v1 , . . . , vr form a set of
linearly independent points in L and so, by Lemma 19, for any a ∈ Fq there is an
r-tuple of integers u = (u1 , . . . , ur ) such that
u1 + u2 R + u3 R2 + · · · + ur Rr−1 = a,
.r
.r
.r
and i=1 |ui | ≤ 12 i=1 5vi 51 . Thus δ(k, q) ≤ 12 i=1 5vi 51 . Now plainly 5vi 51 't
q 1/r , (indeed, as shown in [4], 5vi 51 ≤ r(A(t) + 1)r pn/r , where A(t) is the maximal
absolute value of the coefficients of Φt (x)). Thus δ(k, q) 't q 1/r .
!
7. Proof of Lemma 7(a)
Let m ≥ 3 be a positive integer, A, B ⊆ Fq , A" = A − {0}, a ∈ Fq and N denote
the number of 2m-tuples (x1 , . . . , xm , y1 , . . . ym ) with x1 , x2 ∈ A" , x3 , . . . , xm ∈ A,
yi ∈ B, 1 ≤ i ≤ m, and x1 y1 + · · · + xm ym = a. Let ψ denote the additive character
on Fq , ψ(z) = e2πiT r(z)/p . Then
9 9
9
9
qN = |A" |2 |A|m−2 |B|m +
ψ(λ(x1 y1 + · · · + xm ym − a))
= |A | |A|
" 2
m−2
λ&=0 x1 ,x2 ∈A! x3 ,..,xm ∈A yi ∈B
|B| + Error,
m
(20)
with
Error =
9
λ&=0

ψ(−λa) 
99
x∈A y∈B
m−2 
ψ(λxy)

9 9
x∈A!
y∈B
2
ψ(λxy) .
15
INTEGERS: 11 (2011)
Now, by the Cauchy-Schwarz inequality,
1
1

1
1
12 1/2
19 9
1 9 11 9
1
19
1
9
1
1
1
1
1
1
1
ψ(λxy)11 ≤
ψ(λxy)1 ≤ |B|1/2 
ψ(λxy)1 
1
1
1
1
1
1
1
1x∈A y∈B
1 y∈B x∈A
y∈B x∈A

1
12 1/2
1
9 11 9
1
≤ |B|1/2 
ψ(λxy)1 
= |B|1/2 (q|A|)1/2 .
1
1
1
y∈Fq x∈A
Also,
1
12
1
9 11 9 9
9
9 9
1
1
1 =
ψ(λ(xy))
ψ(λ(x1 y1 − x2 y2 ))
1
1
1
x1 ,x2 ∈A! y1 ,y2 ∈B λ∈Fq
λ∈Fq 1x∈A! y∈B
= q|{(x1 , x2 , y1 , y2 ) : x1 , x2 ∈ A" , y1 , y2 ∈ B, x1 y1 = x2 y2 }| ≤ q|A" |2 |B|,
the latter following since 0 $∈ A" , so x1 y1 = x2 y2 can be written y1 = x−1
1 x2 y2 .
Thus,
1
12
1
1
1
m−2
m−2
m−2 9 1 9 9
1
|Error| ≤ |A| 2 |B| 2 q 2
ψ(λ(xy))11
1
1
λ&=0 1x∈A! y∈B


1
12
19 9
1
9
1
1
m−2
m−2
m−2 

1
= |A| 2 |B| 2 q 2 
ψ(λ(xy))11 − |A" |2 |B|2 
1
1
λ∈Fq 1x∈A! y∈B
@
A
q|A" |2 |B| − |A" |2 |B|2
$
%
m
m
m
= |A| 2 −1 |A" |2 |B| 2 q 2 1 − |B|
,
q
≤ |A|
m−2
2
|B|
m−2
2
q
m−2
2
and we see that the main term in (20) exceeds the error term provided that
$
%
m
m
m
|A| 2 −1 |B| 2 > q 2 1 − |B|
.
q
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